Two ninth-graders, one with three times the mass of the other, attempt a tug-of-role on frictionless ice. Show that the heavier person will gain a speed one-third that of a lighter person.
Forces acting on them must be same but in opposite direction.
The acceleration of the heavier towards the lighter will be one-third(F/3m) as compared to that of lighter (F/m). So their velocities will be unequal after a littl tim hence the rope will become lose and they will collide with each other.
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Forces acting on them must be same but in opposite direction.
The acceleration of the heavier towards the lighter will be one-third(F/3m) as compared to that of lighter (F/m). So their velocities will be unequal after a littl tim hence the rope will become lose and they will collide with each other.
Let's say the lighter is 30kg., the heavier therefore 90kg. They pull, and the lighter person attains 3 m/sec. speed.
(30 x 3) = 90kg/m/sec.
(90/90kg) = 1 m/sec. the heavier attains.
That's 1/3 the speed of the lighter one.
sum of momenta is zero.
So MV+mv=0
0r V/v=--m/M
or V=-v/3
-sign because they travel in opposite directions towards each other
Conservation of momentum...duh!